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STRUCTURE OF NICKEL ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which cannot lead to the correct nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered my Gell-Mann ans Zweig, in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give considerable charge distributions in nucleons for the discovery of nuclear force and structure by applying the laws of electromagnetism. (See my papers of nuclear structure in FUNDAMENTAL PHYSICS CONCEPTS ). Naturally occurring nickel (Ni) is composed of five stable isotopes; Ni-58, Ni-60, Ni-61, Ni-62 and Ni-64 with Ni-58 being the most abundant (68.077% natural abundance). Ni-58 may decay by double beta-plus decay to Fe-58. 26 radioisotopes have been characterized with the most stable being Ni-59 with a half-life of 76,000 years, Ni-63 with a half-life of 100.1 years, and Ni-56 with a half-life of 6.077 days. All of the remaining radioactive isotopes have half-lives that are less than 60 hours and the majority of these have half-lives that are less than 30 seconds. This element also has 1 meta state.The isotopes of nickel range in atomic weight from Ni-48 to Ni-78. Nickel-58 is the most abundant isotope of nickel, making up 68.077% of the natural abundance. Possible sources include electron capture from copper-58 and EC + p from zinc-59. Nickel-59 is a long-lived cosmogenic radionuclide with a half-life of 76,000 years. 59Ni has found many applications in isotope geology. 59Ni has been used to date the terrestrial age of meteorites and to determine abundances of extraterrestrial dust in ice and sediment. ' ' WHY Ni-58, Ni-60, Ni-61 Ni-62 AND Ni-64 ARE STABLE NUCLIDES After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. For understanding why the Ni-58 has a stable structure with respect to the unstable structure of Ni-56 you may read my STRUCTURE OF Ni-56 and Ni-58 . A careful analysis of such a comparison shows that the stable structure of Ni-58 is due to the high symmetry of structure in which the two extra neutrons of opposite spins make two bonds per neutron (one weak and one strong vertical bond) giving enough binding energies to bonds for overcoming the pp and nn repulsions. Whereas the Ni-56 with 28 protons and 28 neutrons has not any extra neutron. It is of interest to note that the protons of the structure of Ni-56 form 8 blank positions able to receive extra neutrons making two bonds per neutron. Moreover for understandind the stable structure of Ni-60 and Ni-64 you can read my STRUCTURE OF Ni-60 AND Ni-64 . ' ' NUCLEAR STRUCTURE OF Ni-57, Ni-59 AND Ni-61 WITH S = -3/2 In the absence of the 2 extra neutrons in the structure of Ni-58 we get the unstable structure of Ni-56 with 27 protons and 27 neutrons. Using the structure of Ni-56 with S =0 we see that in the structure of Ni-57 the deuteron n13p13 changes the spin from S = +1 to S =- 1 because it goes from the +HP1 to -HP2 in order to make bonds with p3 and n4. Since this change gives S =-2 we conclude that the structure of Ni-57 may receive one extra neutron of positive spin. Since this new arrangement of nucleons breaks the high symmetry we see that the one extra neutron is unable to give enough binding energies to pn bonds. Also the Ni-59 with S = -3/2, which has two more extra neutrons of opposite spins cannot give enough binding energies to pn bonds. However the stable structure of Ni-61 is due to the 5 extra neutrons. Each of them makes two bonds per neutron able to overcome the pp and nn repulsions. In other words it has one extra neutron of positive spin and 4 extra neutrons of opposite spins giving S=0. Under this condition the total S =-3/2 of the Ni-61 is given by S = - 2 + 1( +1/2) + 4( 0 ) = -3/2 ' ' NUCLEAR STRUCTURE OF Ni-63, Ni-65, Ni-67 AND Ni-71 After a careful analysis I found that the structure of the above unstable nuclides is based on the structure of Ni-61 with S = -3/2. For example the Ni-65 with S =-5/2 has 2 more extra neutrons of single bonds with negative spins than those of Ni-61 and also 2 more extra neutrons of opposite spins giving S = 0. That is S = -3/2 + 2(-1/2) + 0 = -5/2 ' ' NUCLEAR STRUCTURE OF Ni-55, Ni-53, Ni-51 AND Ni-49 WITH S = -7/2 In the absence of neutrons we see that the structure of the above unstable nuclides is based on a structure which is similar to the structure of Ni-57 with S = -3/2. In this case the deuteron n19p19 of S = +1 which is behind the n2p2 changes the spin from S = +1 to S = -1 because it goes from the +HP1 to -HP2 in order to make bonds with n3p4. Note that this change gives S = -2 as in the case of the movement of the deuteron n13p13 which gives the same S = -2. Under this condition the new structure of Ni-56 will give a spin S = -4. Thus the new structure of Ni-57 will have S = -7/2 because it receives one extra neutron of positive spin. That is S = - 4 + 1(+1/2) = -7/2 ' ' DIAGRAM OF THE NEW STRUCTURE OF Ni-57 WITH S = -7/2 ' p12.........n12' ' -HP6 n11........p11 ' ' n10..........p10' ' +HP5 n29.......p9...........n9 ' ' n26........p8............n8............p28' ' -HP4 p26........n7...........p7.........n28 ' ' p25........n6........... p6............n27 ' ' '' ''+HP3' ' n25.........p5..........n5.........p27 ' ' p19.........p4............n4..........p13' ' -HP2 n19........n3............p3........n13 ' ' n2............p2' ' +HP1 p1...........n1 ' ' ' ' ' Therefore in the absence of neutrons and using the structure of the above diagram with S = -7/2 we get the structure of Ni-53, Ni-51 and Ni-49 with S = -7/2. For example in the structure of Ni-49 with S = -7/2 based on the new structure of Ni-57 with S = -7/2 we have 6 absent neutrons of opposite spins. ' ' NUCLEAR STRUCTURE OF Ni-69, Ni-73, Ni-75 AND Ni-77 When all nucleons of the new structure of Ni-57 change the spins we get the same structure but with a total S = +7/2. In other words in this case we have -HP1, +HP2, -HP3, +HP4, -HP5 and +HP6. Then in such a structure adding extra neutrons with single bonds we get the structure of the above unstable nuclides. For example the Ni-75 with S =+7/2 based on the new structure of Ni-57 with S =+7/2 has 18 extra neutrons of opposite spins. ' ' NUCLEAR STRUCTURE OF Ni-66, Ni-68, Ni-70, Ni-72, Ni-74, ni-76 AND Ni-78 WITH S = 0 The structure of the above unstable nuclides is based on the structure of the stable Ni-64 with S = 0. For example the Ni-78 with S = 0 has 14 more extra neutrons of opposite spins than those of Ni-64 which make single bonds leading to the decay. ' ' NUCLEAR STRUCTURE OF Ni-54, Ni-52, Ni-50 AND Ni-48 WITH S = 0 In the absence of neutrons we see that the structure of the above nuclides is based on the structure of Ni-56 with S = 0. For example in the Ni-48 with S = 0 we have 8 absent neutrons of opposite spins. Category:Fundamental physics concepts